answer:Cash budget

answer:First, we can factor out a 6 from both terms: 6 x^2 - frac{804 x}{7} = 6x(x - frac{134}{7}) Then, we can simplify the expression inside the parentheses: x - frac{134}{7} = frac{7x - 134}{7} Therefore, the fully factored form of the quadratic is: 6x(x - frac{134}{7}) = 6x(frac{7x - 134}{7}) The answer is 6x(x - frac{134}{7})

answer:The integral can be computed as follows: Let's start with the integral int frac{1}{sqrt{4x^2+4x-24}} dx. We observe that 4x^2 + 4x - 24 can be rewritten as 4x^2 + 4x + 1 - 25, which equals (2x + 1)^2 - 5^2. This suggests using the substitution 2x + 1 = 5sectheta, where theta is a trigonometric angle. From this substitution, we get 2dx = 5secthetatantheta dtheta and sqrt{4x^2+4x-24} = sqrt{(2x+1)^2-5^2} = sqrt{25sec^2theta - 25} = 5tantheta. Now, the integral becomes int frac{1}{5tantheta} cdot 5secthetatantheta dtheta = int sectheta dtheta. Upon integrating sectheta, we obtain ln|sectheta + tantheta|. Applying the inverse substitution, we have: int frac{1}{sqrt{4x^2+4x-24}} dx = ln|2x + 1 + sqrt{4x^2+4x-24}| + C Alternatively, we can express this in terms of another constant K: int frac{1}{sqrt{4x^2+4x-24}} dx = ln|2x + 1 + sqrt{4x^2+4x-24}| + K where K = C - ln5. Thus, the integral is completely integrated, and the result reflects the correct use of trigonometric substitution.

answer:We use the substitution {eq}u = ln x {/eq} to simplify the integral. This gives us {eq}du = frac1x dx {/eq}, which we write {eq}dx = x du {/eq}. We also find that {eq}x in [1,2] to u in [ 0, ln 2] {/eq}. Substituting these into the integral, we get {eq}begin{align*} int_{1}^{2}frac{4(ln x)^{3}}{x} dx &= int_0^{ln 2} frac{4u^3}{x} x du &= int_0^{ln 2} 4u^3 du &= left [ u^4 right ]_0^{ln 2} &= (ln 2)^4 &approx 0.23084 end{align*} {/eq} Let {eq}u = ln x {/eq}. Then {eq}du = frac1x dx {/eq}, which we write {eq}dx = x du {/eq}. Also {eq}x in [1,2] to u in [ 0, ln 2] {/eq}. We get {eq}begin{align*} int_{1}^{2}frac{4(ln x)^{3}}{x} dx &= int_0^{ln 2} frac{4u^3}{x} x du &= int_0^{ln 2} 4u^3 du &= left [ u^4 right ]_0^{ln 2} &= (ln 2)^4 &approx 0.23084 end{align*} {/eq}